# capacitor question help

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#1
A capacitor is discharging though a resistor and the time constant is 5.0 s. The time taken for the capacitor to lose half its charge is ?

My answer is 0.14s but the mark scheme said that the answer should be 3.5 s?!
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6 years ago
#2
What is your working out so far?

Hint:
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6 years ago
#3
(Original post by Lamalam)
A capacitor is discharging though a resistor and the time constant is 5.0 s. The time taken for the capacitor to lose half its charge is ?

My answer is 0.14s but the mark scheme said that the answer should be 3.5 s?!
The charge on the capacitor is given by the relationship

ke-t/CR

i.e:

when t = CR (t= 5.0 seconds)

-t/CR = -1

then e-t/CR

e-t/5 = e-1 = 0.3679

i.e. charge or voltage on the capacitor has fallen to approx' 37% of its original value.

The question is asking for the time t such that e-t/5 = 0.5

Can you solve this now?

HINT. Use natural logs
1
#4
(Original post by uberteknik)
The charge on the capacitor is given by the relationship

ke-t/CR

i.e:

when t = CR (t= 5.0 seconds)

-t/CR = -1

then e-t/CR

e-t/5 = e-1 = 0.3679

i.e. charge or voltage on the capacitor has fallen to approx' 37% of its original value.

The question is asking for the time t such that e-t/5 = 0.5

Can you solve this now?

HINT. Use natural logs
Take the natural log of both sides of the equation and rearrange to get t=3.47 s --> 3.5s

thank you !

I calculated the wrong answer at first because I think the time required is actually the HALF LIVE . And then I use half live = ln 2 / RC to get the answer. why the time isn't half lives?
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6 years ago
#5
(Original post by Lamalam)
Take the natural log of both sides of the equation and rearrange to get t=3.47 s --> 3.5s

thank you !

I calculated the wrong answer at first because I think the time required is actually the HALF LIVE . And then I use half live = ln 2 / RC to get the answer. why the time isn't half lives?
You can but you would need to use (RC)ln2 since with half lives, the relationship is

e-lambda x t

However, don't mix the two because the examiner will award marks for the correct use of e-t/CR and none if you used your method which produced the wrong result.
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#6
(Original post by uberteknik)
You can but you would need to use (RC)ln2 since with half lives, the relationship is

e-lambda x t

However, don't mix the two because the examiner will award marks for the correct use of e-t/CR and none if you used your method which produced the wrong result.
silly me ! I rmb the equations wrongly .

e-lambda x t is applicable in this question? I thought this equation can only be used in the radioactivity-related question? if it can be used in the question, what would lambda represents?
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6 years ago
#7
(Original post by Lamalam)
silly me ! I rmb the equations wrongly .

e-lambda x t is applicable in this question? I thought this equation can only be used in the radioactivity-related question? if it can be used in the question, what would lambda represents?
Not really since lambda is the time taken for 50% of the atoms to decay whereas CR is the time for the charge to fall to 37%

As I said, don't mx them up, it only serves to confuse.
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#8
(Original post by uberteknik)
Not really since lambda is the time taken for 50% of the atoms to decay whereas CR is the time for the charge to fall to 37%

As I said, don't mx them up, it only serves to confuse.
If the decay constant is 4.30*10^-4 year^-1, what does it mean? I couldn't associate the 50% with 4.30*10^-4 year^-1. Thank you very much!
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6 years ago
#9
(Original post by Lamalam)
If the decay constant is 4.30*10^-4 year^-1, what does it mean? I couldn't associate the 50% with 4.30*10^-4 year^-1. Thank you very much!
Half life decay is stated as a time. Meaning half of the original quantity will decay within the time period.

Your statement states a quantity of something is decaying at given rate without stating the original quantity. Can you post the original question please?
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#10
(Original post by uberteknik)
Half life decay is stated as a time. Meaning half of the original quantity will decay within the time period.

Your statement states a quantity of something is decaying at given rate without stating the original quantity. Can you post the original question please?

I think I mix up half life and decay constant? Are they the same thing ?

Posted from TSR Mobile
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6 years ago
#11
(Original post by Lamalam)

I think I mix up half life and decay constant? Are they the same thing ?

Posted from TSR Mobile
No. The equations look similar but re not the same.

One refers to an actual physical quantity (rate of change of charge) and the other is the probability of a rate of decay occurring in the stated time.
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#12
(Original post by uberteknik)
No. The equations look similar but re not the same.

One refers to an actual physical quantity (rate of change of charge) and the other is the probability of a rate of decay occurring in the stated time.
but then lambda is not the time taken for 50% of the atoms to decay , but half life does? Thank you for answering my question!
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6 years ago
#13
(Original post by Lamalam)
but then lambda is not the time taken for 50% of the atoms to decay , but half life does? Thank you for answering my question!
Essentially they are the same thing. However probabiity definition is a closer model because radioactive decay cannot be predicted for individual atoms whereas we can indeed measure the decay of half the atoms occuring in a given time for a large bulk of atoms: Statistically half of them will decay within the decay time constant.

So e-t/CR has the same form as e-lambda * t

1/CR = 1/ tor since the product of the dimensions for Farrads and Ohms = seconds

lambda = k/tor = k(1/tor) where k is a property of the atomic element
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